2

我有一个像这样的非数字逻辑对称矩阵

0   0   0   0   0   0   1   1   1   1
0   0   0   0   0   1   0   1   1   1
0   0   0   0   0   1   1   0   1   1
0   0   0   0   0   1   1   0   0   1
0   0   0   0   0   1   1   1   1   0
0   1   1   1   1   0   0   0   0   0
1   0   1   1   1   0   0   0   0   0
1   1   0   0   1   0   0   0   0   0
1   1   1   0   1   0   0   0   0   0
1   1   1   1   0   0   0   0   0   0

我想获得 value == 1 的所有维度（行和列）

结果 = (0,6), (0,7), (0,8), (0,9) (1,5), (1,7), (1,8), (1,9) (2 ,5), (2,6), (2,8), (1,9) (3,5), (3,6), (3,9) (4,5), (4,6), (4,7), (4,8)

在 C# 中使用 ilnumerics 库有更快的方法吗？

编辑：这是我的解决方案

    ILNumerics.ILLogical matrixThreshold;
    ..... Some C# code
    for (int i = 0; i < matrixThreshold.Length; i++)
        for (int j = i + 1; j < matrixThreshold.Length; j++)
            if (matrixThreshold.GetValue(i, j) == 1) Console.Write("({0},{1}){2}", i, j, Environment.NewLine);
4

2 回答 2

2 
ILLogical L = ILMath.rand(10, 12) > 0.5;

>L
Logical [10,12]
    [0]:   0   0   0   1   1   1   0   0   0   0   1   1 
    [1]:   0   1   1   1   0   1   1   1   1   0   1   0 
    [2]:   0   1   1   1   1   1   0   1   1   0   1   1 
    [3]:   0   0   0   0   0   1   0   0   0   0   1   0 
    [4]:   0   0   0   0   0   0   1   0   1   1   0   0 
    [5]:   0   0   1   0   0   1   0   0   1   1   1   1 
    [6]:   1   0   1   0   1   1   0   1   0   1   0   1 
    [7]:   1   1   1   0   1   1   0   1   1   1   0   1 
    [8]:   1   0   0   1   1   1   0   0   1   0   0   0 
    [9]:   1   1   1   1   0   0   0   0   1   1   0   0

成对写出列/行索引：

ILArray<int> C = 1;
ILArray<int> R = ILMath.find(L, 0, C); 
// C now holds the column indices, R the row indices
for (int i = 0; i < C.Length; i++) {
    System.Diagnostics.Debug.WriteLine("({0},{1})", R.GetValue(i), C.GetValue(i)); 
}

这使：

(6,0)
(7,0)
(8,0)
(9,0)
(1,1)
(2,1)
(7,1)
(9,1)
(1,2)
(2,2)
(5,2)
(6,2)
(7,2)
(9,2)
(0,3)
(1,3)
(2,3)
(8,3)
(9,3)
(0,4)
(2,4)
(6,4)
(7,4)
(8,4)
(0,5)
(1,5)
(2,5)
(3,5)
(5,5)
(6,5)
(7,5)
(8,5)
(1,6)
(4,6)
(1,7)
(2,7)
(6,7)
(7,7)
(1,8)
(2,8)
(4,8)
(5,8)
(7,8)
(8,8)
(9,8)
(4,9)
(5,9)
(6,9)
(7,9)
(9,9)
(0,10)
(1,10)
(2,10)
(3,10)
(5,10)
(0,11)
(2,11)
(5,11)
(6,11)
(7,11)
于 2013-10-01T11:31:38.363 回答
-1 
   for(int i=0;i<num1;i++)
   {
    for j=0;j<num2;j++)
    {
     if(a[i][j])
        print("("+i+","+j+")");
      }
   }
于 2013-10-01T09:16:51.413 回答